A paper of Bezrukavnikov (http://arxiv.org/abs/math/0604445) identifies this as $\mathcal{O}(-1) \oplus \mathcal{O}(-1)$ in Example 2.8. What follows is an argument.
The identification of $G$-equivariant vector bundles on $G/B$ with $B$-representations gives us a short exact sequence of $B$-representations: $$0 \rightarrow V_{-2} \rightarrow \mathfrak{b} \rightarrow V_0 \rightarrow 0$$ which gives us a short exact sequence in vector bundles: $$0 \rightarrow \mathcal{O}(-2) \rightarrow \tilde{\mathfrak{sl}_2} \rightarrow \mathcal{O}(0) \rightarrow 0$$ There is no decomposition theorem of $SL_2$-equivariant vector bundles on $\mathbb{P}^1$, since this category is equivalent to $B$-representations and $B$ is solvable. I claim that if we base change to the non-equivariant case, then the short exact sequence becomes: $$0 \rightarrow \mathcal{O}(-2) \rightarrow \mathcal{O}(-1) \oplus \mathcal{O}(-1) \rightarrow \mathcal{O}(0) \rightarrow 0$$
Let $\mathcal{E}$ be the locally free sheaf on $\mathbb{P}^1$ associated to $\tilde{\mathfrak{sl}_2}$. Take the long exact sequence of the short exact sequence of locally free sheaves and one obtains: $$0 \rightarrow H^0(\mathbb{P}^1, \mathcal{E}) \rightarrow H^0(\mathbb{P}^1, \mathcal{O}(0)) \rightarrow H^1(\mathbb{P}^1, \mathcal{O}(-2)) \rightarrow H^1(\mathbb{P}^1, \mathcal{E}) \rightarrow 0.$$ I claim that the first map must be zero. By Borel-Weil-Bott all cohomologies $H^i(\mathbb{P}^1, \mathcal{E})$ are sums of trivial representations (since $H^0(\mathbb{P}^1, \mathcal{O}(0) = H^1(\mathbb{P}^1, \mathcal{O}(-2))$ are trivial); in particular all global sections are $G$-invariant. Above the Borel $$\left(\begin{array}{cc} * & * \\ 0 & * \end{array}\right)$$ take the point in the fiber $$\left(\begin{array}{cc} a & b \\ 0 & -a \end{array}\right)$$ and use the $G$-action to try to move it around. In particular, the action of $$\left(\begin{array}{cc} 1 & t \\ 0 & 1 \end{array}\right)$$ fixes the Borel but moves the section to $$\left(\begin{array}{cc} a & -2at + b \\ 0 & -a \end{array}\right)$$ so the only section which has a chance of invariant under $G$ has $a = 0$ on this fiber (we haven't even considered whether it extends globally). But on this fiber it maps to zero to $\mathcal{O}(0)$ and so by equivariance, if it extends to a global section, this section also maps to zero. So the first map in the short exact sequence is zero, thus the second map is an isomorphism, and the third map is zero. So $\mathcal{E}$ has no cohomology, and by preservation of rank, it must be $\mathcal{O}(-1) \oplus \mathcal{O}(-1)$.
Note: this bundle is as a "twist" of the usual short exact sequence for the tautological bundle on $\mathbb{P}^1$: $$0 \rightarrow \mathcal{O}(-1) \rightarrow \mathcal{O}(0) \oplus \mathcal{O}(0) \rightarrow \mathcal{O}(1) \rightarrow 0.$$